Spatial Information Systems (SIS)Spatial Information Systems (SIS)

COMP 30110COMP 30110

Spatial data structures (1)Spatial data structures (1)

Spatial data structuresSpatial data structures

Data structures that have been defined for manipulating sets Data structures that have been defined for manipulating sets of spatial entitiesof spatial entities

Two aspects:Two aspects:

– storing: space occupied? redundancies?storing: space occupied? redundancies?

– querying, analysing, updating, etc.: time required?querying, analysing, updating, etc.: time required?

In general, it is important to be able to manipulate efficiently, In general, it is important to be able to manipulate efficiently, still maintaining a “low” space complexitystill maintaining a “low” space complexity

Requirement: the stored information must allow for an Requirement: the stored information must allow for an unambiguousunambiguous representation of the subdivision representation of the subdivision

Spatial data structures: evaluationSpatial data structures: evaluation

Spatial data structures can be evaluated in terms of:Spatial data structures can be evaluated in terms of:

– space complexityspace complexity: amount of space (memory) needed for : amount of space (memory) needed for storing storing all information (entities and relations) that is explicitly all information (entities and relations) that is explicitly representedrepresented

– time complexitytime complexity of the algorithms for calculating relations of the algorithms for calculating relations that that are not explicitly representedare not explicitly represented

Constant and variable relationsConstant and variable relations

In a generic plane subdivision, a relation is:In a generic plane subdivision, a relation is:

- - constantconstant if it involves a constant number of entities if it involves a constant number of entities

(e.g.: EV, EE, EF involve two entities each)(e.g.: EV, EE, EF involve two entities each)

- - variablevariable if it involves a variable number of entities if it involves a variable number of entities

(e.g.: VV, VE, VF, FV, FE, FF involve a different number of entities (e.g.: VV, VE, VF, FV, FE, FF involve a different number of entities depending on the different cases)depending on the different cases)

Pjj

Pii

e

fjj

fiieii ejj

f f4

f3

f2

f1P1P2

P3 P4

e1

e2

e3

e4

P1P2

P3

P e1e2

e3f1

f2

f3

Topology/ConnectivityTopology/Connectivity

•Generic sets of entities: any possible topological relationsGeneric sets of entities: any possible topological relations

• Overlayed sets of entities (plane subdivisions): only Overlayed sets of entities (plane subdivisions): only meetmeet and and disjointdisjoint

•MeetMeet: topological relation that defines : topological relation that defines connectivityconnectivity between entities in a plane subdivisionbetween entities in a plane subdivision

•Entities of different dimension are “connected” in Entities of different dimension are “connected” in different ways: relations (vertex-, edge-, face-based) seen different ways: relations (vertex-, edge-, face-based) seen beforebefore

Spaghetti data structureSpaghetti data structure

• Spaghetti data structureSpaghetti data structure: represents sets of points, : represents sets of points, lines and polygonslines and polygons

• Can be used for both generic sets of entities and Can be used for both generic sets of entities and overlayed sets (plane subdivisions)overlayed sets (plane subdivisions)

• The geometry of any spatial entity is described The geometry of any spatial entity is described independently of other entitiesindependently of other entities

• No topology/connectivity information is recordedNo topology/connectivity information is recorded

Spaghetti data structure: exampleSpaghetti data structure: example

Example:Example: points, lines and polygons are stored separately points, lines and polygons are stored separately

Polygon 1 Polygon 2

X1,Y1

X2,Y2

X3,Y3

X4,Y4

X5,Y5

X6,Y6

X7,Y7

X1,Y1

X8,Y8

X9,Y9

X10,Y10

X11,Y11

X3,Y3

X2,Y2

X1,Y1

X8,Y8

1 2X5,Y5

X4,Y4

X3,Y3X11,Y11

X1,Y1

X2,Y2

X7,Y7X6,Y6

X10,Y10

X9,Y9

X8,Y8

For each polygon, we store a (ordered) list of coordinates of points on its For each polygon, we store a (ordered) list of coordinates of points on its boundary (this is not the same as FV relation!)boundary (this is not the same as FV relation!)

Another data structureAnother data structure

Polygon 1

Polygon 2

P1

P2

P3

P4

P5

P6

P7

P1

P8

P9

P10

P11

P3

P2

P1

P8

1 2P5

P4

P3P11

P1

P2

P7P6

P10

P9

P8

Points and polygons are related: Points and polygons are related: for each polygon, we store a for each polygon, we store a (ordered) list of (ordered) list of pointspoints (not (not coordinates) on its coordinates) on its boundary: boundary: FV relationFV relation

ID Coord.

P1

P2

P3

P4

P5

P6

P7

P8

P9

P10

P11

X1,Y1

X2,Y2

X3,Y3

X4,Y4

X5,Y5

X6,Y6

X7,Y7

X8,Y8

X9,Y9

X10,Y10

X11,Y11

Spaghetti data structure: pros & consSpaghetti data structure: pros & cons

•Advantages: Advantages:

– simplicitysimplicity

– easy insertions of new entities (all entities are independent)easy insertions of new entities (all entities are independent)

•Cons: Cons:

– inefficient for topological queriesinefficient for topological queries

– redundancies (and consequently, possible inconsistencies!)redundancies (and consequently, possible inconsistencies!)

Spaghetti data structure: remarksSpaghetti data structure: remarks

Example:Example:

NOTE 1NOTE 1:: coordinates of points along common boundary are recorded coordinates of points along common boundary are recorded twice!twice!

Redundancy: if we update coordinates of a point, we need to update Redundancy: if we update coordinates of a point, we need to update them everywhere!them everywhere!

Polygon 1 Polygon 2

X1,Y1

X2,Y2

X3,Y3

X4,Y4

X5,Y5

X6,Y6

X7,Y7

X1,Y1

X8,Y8

X9,Y9

X10,Y10

X11,Y11

X3,Y3

X2,Y2

X1,Y1

X8,Y8

1 2X5,Y5

X4,Y4

X3,Y3X11,Y11

X1,Y1

X2,Y2

X7,Y7X6,Y6

X10,Y10

X9,Y9

X8,Y8

Spaghetti data structure: remarks (cont.d)Spaghetti data structure: remarks (cont.d)

Example:Example:

NOTE 2NOTE 2:: no easy way of solving queries such as: no easy way of solving queries such as: “Do Polygon 1 and 2 “Do Polygon 1 and 2 share a common bounding line?”share a common bounding line?”

Need to analyse all coordinates of points of Polygon 1 and compare Need to analyse all coordinates of points of Polygon 1 and compare with those of Polygon 2 and see if two consecutive pairs are the same: with those of Polygon 2 and see if two consecutive pairs are the same: inefficient!!inefficient!!

Polygon 1 Polygon 2

X1,Y1

X2,Y2

X3,Y3

X4,Y4

X5,Y5

X6,Y6

X7,Y7

X1,Y1

X8,Y8

X9,Y9

X10,Y10

X11,Y11

X3,Y3

X2,Y2

X1,Y1

X8,Y8

1 2X5,Y5

X4,Y4

X3,Y3X11,Y11

X1,Y1

X2,Y2

X7,Y7X6,Y6

X10,Y10

X9,Y9

X8,Y8

Topological data structures for plane subdivisionsTopological data structures for plane subdivisions

• Incorporate “connectivity” among spatial objects by Incorporate “connectivity” among spatial objects by storing explicitly a subset of the relations seen beforestoring explicitly a subset of the relations seen before

• Two types of points: Two types of points: – points that are endpoints of a line (vertices)points that are endpoints of a line (vertices)

– points that define the shape (geometry) of a linepoints that define the shape (geometry) of a line

• For polygons/lines: For polygons/lines: – adjacency information (e.g., encode polygons on each side of a adjacency information (e.g., encode polygons on each side of a boundary line – left and right polygon with respect to the boundary line – left and right polygon with respect to the direction of the line: EF relation)direction of the line: EF relation)

Topological data structures: motivationsTopological data structures: motivations

Storing connectivity information explicitly allows for more Storing connectivity information explicitly allows for more efficient spatial queriesefficient spatial queries

ExampleExample: :

if we store relation if we store relation FEFE explicitely (i.e., for each polygon we explicitely (i.e., for each polygon we

store a list of IDs of edges bounding it), the query:store a list of IDs of edges bounding it), the query:

““do Polygon 1 and 2 share a common bounding line?”do Polygon 1 and 2 share a common bounding line?”

only requires checking whether the two lists contain any common IDsonly requires checking whether the two lists contain any common IDs

Polygon 1

Polygon 2

e1

e2

e3

e4

e5

e6

e7

e8

e9

e10

e11

e12

e2

e1

1 2e4

e3 e11

e7

e1

e6e5

e10

e9

e8

e2

e12

Topological data structures: motivations Topological data structures: motivations (cont.d)(cont.d)

Topology/connectivity: important criterion to establish the Topology/connectivity: important criterion to establish the correctness (integrity, consistency) of geographic databasescorrectness (integrity, consistency) of geographic databases

ExamplesExamples::

– neighbouring countries must share a boundaryneighbouring countries must share a boundary

– seas and lakes cannot overlapseas and lakes cannot overlap

Doubly-Connected Edge List (DCEL) structureDoubly-Connected Edge List (DCEL) structure

(Preparata and Shamos, 1985)(Preparata and Shamos, 1985)

• DCELDCEL structure stores: structure stores:– the three sets of entities V, E, Fthe three sets of entities V, E, F

– the three edge-based relations EV, EE, EF (all constant)the three edge-based relations EV, EE, EF (all constant)

– the two partial relations: FE* and VE* the two partial relations: FE* and VE*

FE*FE*: : associates a face associates a face f f with with one of one of

the edges bounding the edges bounding ff

VE*VE*: : associates a vertex associates a vertex v v with onewith one

of the edges incident in of the edges incident in vvV F

EEF

FE**EV

VE**

EE

Example (1)Example (1)

P9

P8

P3

f1 f2e4

e3 e11

e7

e1

e6e5

e10

e9

e8

e2

e12

P5

P4P11

P1

P2

P7P6

P10f0

Entities

V {P1,P2,….., P11}

E {e1, e2,….., e12}

F {f0,f1,f2}

Partial relations

VE* FE*

P1 e1 f0 e3

P2 e2 f1 e3

P3 e3 f2 e1

P4 e4

P5 e5

P6 e6

P7 e7

P8 e8

P9 e9

P10 e10

P11 e11

Example (2)Example (2) Entities

V P1,P2,….., P11

E e1, e2,….., e12

F f0,f1,f2

Edge - based relations

EV EF EE

e1 P1,P2 f1,f2 e7,e2

e2 P2,P3 f1,f2 e1,e12

e3 P3,P4 f1,f0 e2,e4

e4 P4,P5 f1,f0 e3,e5

e5 P5,P6 f1,f0 e4,e6

e6 P6,P7 f1,f0 e5,e7

e7 P7,P1 f1,f0 e6,e8

e8 P1,P8 f2,f0 e1,e9

e9 P8,P9 f2,f0 e8,e10

e10 P9,P10 f2,f0 e9,e11

e11 P10,P11 f2,f0 e10,e12

e12 P11,P3 f2,f0 e11,e3

P9

P8

P3

f1 f2e4

e3 e11

e7

e1

e6e5

e10

e9

e8

e2

e12

P5

P4P11

P1

P2

P7P6

P10f0

DCEL: space complexityDCEL: space complexity

For every edge:For every edge:

3 constant relations are stored (involving 2 entities): 63 constant relations are stored (involving 2 entities): 6ee

For every face:For every face:

1 relation involving one entity: 1 relation involving one entity: ff

For every vertex:For every vertex:

1 relation involving one entity: 1 relation involving one entity: nn

Space required to represent relations:Space required to represent relations: 66e + f + ne + f + n

For each vertex we also store the two geometric coordinates: 2For each vertex we also store the two geometric coordinates: 2nn

Retrieving missing relations: FERetrieving missing relations: FE

Calculating complete relation FE:Calculating complete relation FE: Obtained by combining FE Obtained by combining FE* * and and EEEE

For example, given a face For example, given a face f1f1, we find the first bounding edge , we find the first bounding edge e3e3 using FEusing FE**. Then using EE we find the successor . Then using EE we find the successor e4e4 ofof e3e3 (in (in counter-closkwise order) on the boundary ofcounter-closkwise order) on the boundary of f1: f1: if if e3e3 is oriented in is oriented in such a way that such a way that f1f1 is on its left hand side, then is on its left hand side, then e4e4 is the second of is the second of the two edges associated with the two edges associated with e3e3 through EEthrough EE

P9

P8

P3

f1 f2e4

e3 e11

e7

e1

e6e5

e10

e9

e8

e2

e12

P5

P4P11

P1

P2

P7P6

P10f0

Retrieving missing relations: FE (cont.d)Retrieving missing relations: FE (cont.d)

We apply the same method (2We apply the same method (2ndnd element of EE) to obtain all other element of EE) to obtain all other edges on the boundary of edges on the boundary of f1f1, until we reach , until we reach e3e3 again. again.

P9

P8

P3

f1 f2e4

e3 e11

e7

e1

e6e5

e10

e9

e8

e2

e12

P5

P4P11

P1

P2

P7P6

P10f0

DCEL: FV and FFDCEL: FV and FF

FV can be obtained by combining FE and EV: for each bounding FV can be obtained by combining FE and EV: for each bounding edge edge ee of a given face of a given face f f (obtained with FE), we consider its (obtained with FE), we consider its endpoints using EVendpoints using EV

FF can be obtained by combining FE and EF: for each bounding FF can be obtained by combining FE and EF: for each bounding edge edge ee of a given face of a given face f f (obtained with FE), we consider the other (obtained with FE), we consider the other face face f’ f’ obtained by using EFobtained by using EF

DCEL: VE, VV, VFDCEL: VE, VV, VF

Exercise!!Exercise!!

For next weekFor next week